Question: What is the area of the region between the graphs of $f(x)=x^2+2x+2$ and $g(x)=2x^2+3x-4$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{125}{6}$ (Choice B) B $\dfrac{22}{3}$ (Choice C) C $\dfrac{27}{2}$ (Choice D) D $5$
Solution: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${1}$ ${2}$ ${\llap{-}2}$ ${\llap{-}3}$ ${2}$ ${4}$ ${6}$ ${8}$ ${10}$ ${\llap{-}4}$ $f$ $g$ $y$ $x$ From the graph, it appears that $f(x)\ge g(x)$ between the points where the graphs intersect. From this we are looking to evaluate: $ \int_{a}^{b}\left( f(x)-g(x) \right)\,dx$ where $a$ and $b$ are the $x$ -coordinates of the points of intersection. Finding the $x$ -coordinates of the intersection points We can find the $x$ -coordinate of each point of intersection by setting the functions equal to each other and solving the resulting equation. $\begin{array}{rl} f(x)&=g(x) \\\\ x^2+2x+2 &=2x^2+3x-4\\\\ 0&=x^2+x-6 \\\\ 0&=(x+3)(x-2) \end{array}$ The graphs intersect where $x=-3$ and $x=2$. Setting up the definite integral Thus, the area of the shaded region pictured above is given by: $\begin{aligned} &\phantom{=} \int_{-3}^{2}\left(x^2+2x+2 -(2x^2+3x-4)\right)\,dx \\\\ &= \int_{-3}^{2} \left( -x^2-x+6 \right) \,dx \end{aligned}$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{-3}^{2} \left( -x^2-x+6 \right) \,dx \\\\ &= -\dfrac{x^3}{3}-\dfrac{x^2}{2}+6x~\Bigg|_{-3}^{2}\\\\ &= \left( -\dfrac{8}{3}-2+12 \right) - \left( 9-\dfrac{9}{2}-18 \right) \\\\ &= \dfrac{125}{6} \end{aligned}$ Answer The area is $\dfrac{125}{6}$ square units.